💡First, I recognize that the intersection of sets ( A ) and ( B ) is given as ( {2, 3} ). This means that both elements ( 2 ) and ( 3 ) must be present in both sets ( A ) and ( B ).Looking at set ( A = {1, 2, 3} ), it already contains the elements ( 2 ) and ( 3 ). Therefore, set ( B = {m, 3, 6} ) must also contain these elements for the intersection to be ( {2, 3} ).Since ( 3 ) is already in set ( B ), the missing element in ( B ) to satisfy the intersection is ( 2 ). Therefore, ( m ) must be ( 2 ).By assigning ( m = 2 ), set ( B ) becomes ( {2, 3, 6} ), which ensures that the intersection ( A cap B = {2, 3} ) holds true.

💡First, I need to find the semiperimeter of the triangle using the given side lengths.Next, I'll calculate the area of the triangle using Heron's formula.Once I have the area, I can use the relationship between the area, semiperimeter, and inradius to find the radius of the inscribed circle.Finally, I'll solve for the inradius using the derived formula.

💡First, I need to determine the distance the wheel covers in one revolution, which is the circumference of the wheel.Given that the wheel covers 91.77 meters in 100 revolutions, I can find the circumference by dividing the total distance by the number of revolutions.Once I have the circumference, I can use the formula for the circumference of a circle, ( C = 2pi r ), to solve for the radius ( r ).After calculating the radius in meters, I will convert it to centimeters by multiplying by 100, since there are 100 centimeters in a meter.

💡First, I need to determine how much Stephanie has already paid for each bill and how much remains unpaid.For the electricity bill, it's already paid in full, so there's nothing left to pay.Next, for the gas bill, she has paid three-quarters of the 40 bill, which is 30. She then makes an additional payment of 5, bringing the total paid to 35. This leaves 5 remaining on the gas bill.For the water bill, she has paid half of the 40 bill, which is 20. Therefore, 20 remains to be paid.Lastly, for the internet bill, she has made 4 payments of 5, totaling 20. This leaves 5 remaining on the internet bill.Finally, I'll add up the remaining amounts for each bill to find the total amount Stephanie still needs to pay.

💡To calculate the residue of (-963 + 100) modulo (35), I first simplify the expression by performing the addition: (-963 + 100 = -863). Next, I need to find (-863 pmod{35}). This involves determining how many times (35) fits into (-863) and finding the remainder. Dividing (-863) by (35) gives approximately (-24.657), and rounding down to the nearest integer quotient gives (-25). Multiplying (35) by (-25) results in (-875). Subtracting this from (-863) gives the remainder: (-863 - (-875) = 12). Since (12) is within the range (0) to (34), the residue of (-963 + 100) modulo (35) is (12).

💡Okay, so I have this problem here where I need to find the positive value of base ( b ) such that the base five number ( 43_5 ) is equal to the base ( b ) number ( 121_b ). Hmm, let me try to break this down step by step.First, I remember that when numbers are written in a certain base, each digit represents a power of that base. For example, in base five, the number ( 43_5 ) has two digits: 4 and 3. The rightmost digit is the units place, which is ( 5^0 ), and the next digit to the left is the fives place, which is ( 5^1 ). So, to convert ( 43_5 ) to decimal (base ten), I can calculate it as:[4 times 5^1 + 3 times 5^0]Calculating that, ( 5^1 ) is 5, so ( 4 times 5 = 20 ). Then, ( 5^0 ) is 1, so ( 3 times 1 = 3 ). Adding those together, ( 20 + 3 = 23 ). So, ( 43_5 ) is equal to 23 in decimal.Now, I need to figure out what base ( b ) makes the number ( 121_b ) equal to 23 in decimal. Let me recall how to convert a number from base ( b ) to decimal. The number ( 121_b ) has three digits: 1, 2, and 1. Starting from the right, the digits represent ( b^0 ), ( b^1 ), and ( b^2 ) respectively. So, converting ( 121_b ) to decimal would be:[1 times b^2 + 2 times b^1 + 1 times b^0]Simplifying that, it's ( b^2 + 2b + 1 ). So, I have the equation:[b^2 + 2b + 1 = 23]Now, I need to solve this equation for ( b ). Let me subtract 23 from both sides to set the equation to zero:[b^2 + 2b + 1 - 23 = 0][b^2 + 2b - 22 = 0]This is a quadratic equation in the form ( ax^2 + bx + c = 0 ). To solve for ( b ), I can use the quadratic formula:[b = frac{-B pm sqrt{B^2 - 4AC}}{2A}]In this equation, ( A = 1 ), ( B = 2 ), and ( C = -22 ). Plugging these values into the quadratic formula:[b = frac{-2 pm sqrt{(2)^2 - 4(1)(-22)}}{2(1)}][b = frac{-2 pm sqrt{4 + 88}}{2}][b = frac{-2 pm sqrt{92}}{2}]Simplifying ( sqrt{92} ), I know that 92 can be broken down into 4 times 23, so:[sqrt{92} = sqrt{4 times 23} = 2sqrt{23}]So, substituting back into the equation:[b = frac{-2 pm 2sqrt{23}}{2}]I can factor out a 2 in the numerator:[b = frac{2(-1 pm sqrt{23})}{2}]Canceling out the 2:[b = -1 pm sqrt{23}]Now, I have two possible solutions:1. ( b = -1 + sqrt{23} )2. ( b = -1 - sqrt{23} )Since a base cannot be negative, I discard the second solution because ( -1 - sqrt{23} ) is definitely negative. So, I'm left with:[b = -1 + sqrt{23}]Calculating ( sqrt{23} ) approximately, since ( 4^2 = 16 ) and ( 5^2 = 25 ), ( sqrt{23} ) is somewhere between 4 and 5. Let's approximate it:[sqrt{23} approx 4.796]So,[b approx -1 + 4.796 = 3.796]Hmm, that's approximately 3.796, which is not an integer. But bases are typically integers because they represent the number of distinct digits in a numbering system. For example, base 5 has digits 0-4, base 10 has digits 0-9, etc. If the base were a non-integer, it wouldn't make much sense in standard positional numeral systems.Wait, maybe I made a mistake somewhere. Let me double-check my calculations.Starting again, I converted ( 43_5 ) correctly to decimal:[4 times 5 + 3 = 20 + 3 = 23]That seems right. Then, converting ( 121_b ) to decimal:[1 times b^2 + 2 times b + 1 = b^2 + 2b + 1]Setting that equal to 23:[b^2 + 2b + 1 = 23][b^2 + 2b - 22 = 0]Quadratic formula:[b = frac{-2 pm sqrt{4 + 88}}{2} = frac{-2 pm sqrt{92}}{2}]Which simplifies to:[b = frac{-2 pm 2sqrt{23}}{2} = -1 pm sqrt{23}]So, the calculations seem correct. But since ( b ) must be a positive integer, and ( sqrt{23} ) is approximately 4.796, ( b ) would be approximately 3.796, which is not an integer. That's a problem because bases are integers.Wait, maybe I made a mistake in setting up the equation. Let me check that again.The number ( 121_b ) is equal to 23 in decimal. So, ( 1 times b^2 + 2 times b + 1 = 23 ). That seems correct.Alternatively, perhaps I need to consider that the digits in ( 121_b ) must be valid in base ( b ). That is, each digit must be less than ( b ). So, in ( 121_b ), the digits are 1, 2, and 1. The highest digit is 2, so ( b ) must be greater than 2. But that doesn't necessarily solve the problem because ( b ) is approximately 3.796, which is greater than 2, but it's not an integer.Wait, maybe I should try plugging in integer values for ( b ) near 3.796 to see if any of them satisfy the equation ( b^2 + 2b + 1 = 23 ).Let's try ( b = 4 ):[4^2 + 2 times 4 + 1 = 16 + 8 + 1 = 25]25 is greater than 23.Next, ( b = 3 ):[3^2 + 2 times 3 + 1 = 9 + 6 + 1 = 16]16 is less than 23.So, between ( b = 3 ) and ( b = 4 ), the value of ( b^2 + 2b + 1 ) goes from 16 to 25. Since 23 is between 16 and 25, ( b ) must be between 3 and 4. But since ( b ) must be an integer, there's no integer ( b ) that satisfies the equation exactly.Wait, that can't be right because the problem states that ( 43_5 ) is equal to ( 121_b ), so there must be a solution. Maybe I made a mistake in the initial conversion.Let me double-check the conversion of ( 43_5 ) to decimal. ( 4 times 5^1 + 3 times 5^0 = 20 + 3 = 23 ). That seems correct.Wait, perhaps the problem is that I'm assuming ( b ) must be an integer, but maybe it's allowed to be a non-integer? But in standard numeral systems, bases are integers. So, that seems unlikely.Alternatively, maybe I misread the problem. Let me check again: "The base five number ( 43_5 ) is equal to the base ( b ) number ( 121_b ). What is the positive value of ( b )?" So, it's definitely asking for a positive value of ( b ), but it doesn't specify that ( b ) has to be an integer. However, in standard contexts, bases are integers, so perhaps the problem expects an integer solution, and I might have made a mistake in my calculations.Wait, let me try another approach. Maybe I can factor the quadratic equation ( b^2 + 2b - 22 = 0 ). Let's see if it factors nicely.Looking for two numbers that multiply to -22 and add up to 2. Hmm, 11 and -2: 11 * (-2) = -22, and 11 + (-2) = 9. Not 2.How about 11 and -2? No, same as above.Wait, maybe it doesn't factor nicely, which is why I had to use the quadratic formula. So, perhaps the solution is indeed ( b = -1 + sqrt{23} ), which is approximately 3.796, but since bases are integers, maybe the problem expects me to round it or consider the closest integer.But rounding 3.796 would give me 4, but when I plug ( b = 4 ) into ( 121_b ), I get 25, which is not 23. So, that doesn't work.Wait, maybe I made a mistake in setting up the equation. Let me check again.( 121_b ) in decimal is ( 1 times b^2 + 2 times b + 1 ). That's correct.Set equal to 23:[b^2 + 2b + 1 = 23][b^2 + 2b - 22 = 0]Yes, that's correct.Wait, maybe I can try completing the square instead of using the quadratic formula.Starting with ( b^2 + 2b - 22 = 0 ).Move the constant term to the other side:[b^2 + 2b = 22]To complete the square, take half of the coefficient of ( b ), which is 1, square it to get 1, and add it to both sides:[b^2 + 2b + 1 = 22 + 1][(b + 1)^2 = 23]Taking the square root of both sides:[b + 1 = pm sqrt{23}]So,[b = -1 pm sqrt{23}]Again, same result. So, ( b = -1 + sqrt{23} ) or ( b = -1 - sqrt{23} ). Since ( b ) must be positive, ( b = -1 + sqrt{23} approx 3.796 ).But as I thought earlier, this is not an integer, which is confusing because the problem seems to imply that such a base exists. Maybe I need to reconsider my approach.Wait, perhaps I misread the original numbers. Let me check again: ( 43_5 ) and ( 121_b ). Yes, that's correct.Alternatively, maybe the problem is in a different base system where non-integer bases are allowed, but that's highly unconventional. Most numeral systems use integer bases.Wait, another thought: perhaps I made a mistake in converting ( 43_5 ) to decimal. Let me double-check that.( 43_5 ) is 4 in the fives place and 3 in the units place. So, 4 * 5 + 3 * 1 = 20 + 3 = 23. That's correct.Hmm, I'm stuck. The math seems to lead me to a non-integer base, which doesn't make sense in standard contexts. Maybe the problem has a typo, or perhaps I'm missing something.Wait, perhaps the number ( 121_b ) is not a three-digit number but a two-digit number? Let me check the original problem again: "The base five number ( 43_5 ) is equal to the base ( b ) number ( 121_b )." So, it's definitely a three-digit number in base ( b ).Wait, maybe I should consider that in some numeral systems, the digits can be equal to the base, but actually, no, in standard positional numeral systems, each digit must be less than the base. So, in ( 121_b ), the digits are 1, 2, and 1, so the base must be greater than 2, which it is in this case.But again, the base is approximately 3.796, which is greater than 2, but not an integer.Wait, perhaps the problem is designed to have a non-integer base, but that's unusual. Alternatively, maybe I made a mistake in the initial setup.Wait, let me try another approach. Maybe I can express ( 121_b ) as ( (b + 1)^2 ), since ( 121_b = 1 times b^2 + 2 times b + 1 = (b + 1)^2 ). So, ( (b + 1)^2 = 23 ).Taking the square root of both sides:[b + 1 = sqrt{23}][b = sqrt{23} - 1]Which is the same result as before, approximately 3.796.So, unless the problem allows for a non-integer base, which is unconventional, I might have to conclude that there's no integer solution, but the problem states that such a base exists. Therefore, perhaps I made a mistake in my calculations.Wait, let me try plugging ( b = 4 ) into ( 121_b ):[1 times 4^2 + 2 times 4 + 1 = 16 + 8 + 1 = 25]Which is 25, not 23.What about ( b = 3 ):[1 times 3^2 + 2 times 3 + 1 = 9 + 6 + 1 = 16]Which is 16, not 23.So, between ( b = 3 ) and ( b = 4 ), the value of ( 121_b ) goes from 16 to 25. Since 23 is between 16 and 25, ( b ) must be between 3 and 4, but not an integer.Wait, maybe the problem is designed to have a non-integer base, but that's highly unusual. Alternatively, perhaps I misread the original numbers.Wait, another thought: maybe the number ( 43_5 ) is not in base five, but in another base? No, the problem clearly states base five.Alternatively, perhaps the number ( 121_b ) is in a different form. Wait, no, it's a standard three-digit number in base ( b ).Hmm, I'm stuck. The math leads me to a non-integer base, which doesn't make sense in standard contexts. Maybe the problem is designed to have a non-integer base, but that's unconventional. Alternatively, perhaps I made a mistake in my calculations.Wait, let me try solving the quadratic equation again:[b^2 + 2b - 22 = 0]Using the quadratic formula:[b = frac{-2 pm sqrt{(2)^2 - 4(1)(-22)}}{2(1)} = frac{-2 pm sqrt{4 + 88}}{2} = frac{-2 pm sqrt{92}}{2}]Simplifying ( sqrt{92} ):[sqrt{92} = sqrt{4 times 23} = 2sqrt{23}]So,[b = frac{-2 pm 2sqrt{23}}{2} = -1 pm sqrt{23}]So, ( b = -1 + sqrt{23} ) or ( b = -1 - sqrt{23} ). Since ( b ) must be positive, ( b = -1 + sqrt{23} approx 3.796 ).But since bases are integers, maybe the problem expects me to round it to the nearest integer, which would be 4, even though ( 121_4 = 25 ), which is not 23. Alternatively, perhaps the problem is designed to have a non-integer base, but that's unconventional.Wait, maybe I made a mistake in the initial conversion of ( 43_5 ). Let me check again:[43_5 = 4 times 5^1 + 3 times 5^0 = 20 + 3 = 23]That's correct.Alternatively, perhaps the number ( 121_b ) is not a three-digit number but a two-digit number. Let me check the original problem again: "The base five number ( 43_5 ) is equal to the base ( b ) number ( 121_b )." So, it's definitely a three-digit number in base ( b ).Wait, another thought: maybe the problem is in a different base system where digits can be equal to or greater than the base, but that's not standard. In standard systems, each digit must be less than the base.Wait, perhaps I should consider that ( b ) could be a fraction. Let me explore that possibility.If ( b ) is a fraction, say ( b = frac{m}{n} ) where ( m ) and ( n ) are integers, then ( b^2 + 2b + 1 = 23 ) would still hold. But solving for ( b ) as a fraction might complicate things, and it's not clear if that's intended.Alternatively, maybe the problem is designed to have a non-integer base, but that's highly unconventional. I think I might have to conclude that the problem, as stated, leads to a non-integer base, which is unusual but mathematically valid.Therefore, the positive value of ( b ) is ( sqrt{23} - 1 ), which is approximately 3.796.But since the problem asks for the positive value of ( b ), and it doesn't specify that ( b ) must be an integer, I think that's the answer.Wait, but in the initial problem, the user provided a solution where ( b = 4 ), which seems incorrect because ( 121_4 = 25 ), not 23. Maybe the user made a mistake in their solution.Alternatively, perhaps I made a mistake in my calculations. Let me try one more time.Starting with ( 43_5 = 23 ) in decimal.Then, ( 121_b = b^2 + 2b + 1 = 23 ).So,[b^2 + 2b + 1 = 23][b^2 + 2b - 22 = 0]Using the quadratic formula:[b = frac{-2 pm sqrt{4 + 88}}{2} = frac{-2 pm sqrt{92}}{2} = frac{-2 pm 2sqrt{23}}{2} = -1 pm sqrt{23}]So, ( b = -1 + sqrt{23} approx 3.796 ) or ( b = -1 - sqrt{23} approx -5.796 ). Since we're looking for the positive value, ( b approx 3.796 ).Therefore, the positive value of ( b ) is ( sqrt{23} - 1 ), which is approximately 3.796.But since the problem didn't specify that ( b ) must be an integer, I think that's the correct answer, even though it's not an integer. However, in standard numeral systems, bases are integers, so this is a bit confusing.Wait, perhaps the problem is designed to have a non-integer base, but that's unconventional. Alternatively, maybe I made a mistake in my calculations.Wait, let me try plugging ( b = sqrt{23} - 1 ) back into ( 121_b ):[1 times (sqrt{23} - 1)^2 + 2 times (sqrt{23} - 1) + 1]Expanding ( (sqrt{23} - 1)^2 ):[(sqrt{23})^2 - 2sqrt{23} + 1 = 23 - 2sqrt{23} + 1 = 24 - 2sqrt{23}]Then, adding ( 2 times (sqrt{23} - 1) ):[24 - 2sqrt{23} + 2sqrt{23} - 2 + 1 = 24 - 2 + 1 = 23]Yes, it checks out. So, ( b = sqrt{23} - 1 ) is indeed the correct solution, even though it's not an integer.Therefore, the positive value of ( b ) is ( sqrt{23} - 1 ), which is approximately 3.796.